1. **State the problem:** Calculate the definite integral $$\int_{\frac{3\pi}{2}}^{2\pi} 35 \sin^4 x \cos^3 x \, dx$$. 2. **Recall the formula and approach:** We want to integrate a product of powers of sine and cosine. A common strategy is to use substitution and trigonometric identities. 3. **Rewrite the integral:** $$\int_{\frac{3\pi}{2}}^{2\pi} 35 \sin^4 x \cos^3 x \, dx = 35 \int_{\frac{3\pi}{2}}^{2\pi} \sin^4 x \cos^3 x \, dx$$ 4. **Use substitution:** Let $$u = \sin x$$, then $$du = \cos x \, dx$$. Rewrite $$\cos^3 x = \cos^2 x \cos x = (1 - \sin^2 x) \cos x = (1 - u^2) \cos x$$. So the integral becomes: $$35 \int_{x=\frac{3\pi}{2}}^{2\pi} u^4 (1 - u^2) \cos x \, dx = 35 \int_{u=\sin(\frac{3\pi}{2})}^{\sin(2\pi)} u^4 (1 - u^2) \, du$$ 5. **Evaluate the new limits:** $$\sin\left(\frac{3\pi}{2}\right) = -1$$ $$\sin(2\pi) = 0$$ So the integral is: $$35 \int_{-1}^{0} u^4 (1 - u^2) \, du = 35 \int_{-1}^{0} (u^4 - u^6) \, du$$ 6. **Integrate term-by-term:** $$\int u^4 \, du = \frac{u^5}{5}$$ $$\int u^6 \, du = \frac{u^7}{7}$$ So: $$35 \left[ \frac{u^5}{5} - \frac{u^7}{7} \right]_{-1}^{0}$$ 7. **Evaluate at the bounds:** At $$u=0$$: $$\frac{0^5}{5} - \frac{0^7}{7} = 0$$ At $$u=-1$$: $$\frac{(-1)^5}{5} - \frac{(-1)^7}{7} = \frac{-1}{5} - \frac{-1}{7} = -\frac{1}{5} + \frac{1}{7} = -\frac{7}{35} + \frac{5}{35} = -\frac{2}{35}$$ 8. **Subtract:** $$35 \left(0 - \left(-\frac{2}{35}\right)\right) = 35 \times \frac{2}{35} = 2$$ **Final answer:** $$\boxed{2}$$